# Ngô Quốc Anh

## April 16, 2010

### Kelvin transform: Biharmonic

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 0:36

In this topic, we consider the Kelvin transform for Laplaction operators. Precisely, what we get is the following

$\displaystyle\Delta \left( {\frac {1} {{{{\left| x \right|}^{n - 2}}}}u\left( {\frac {x} {{{{\left| x \right|}^2}}}} \right)} \right){\text{ }} = \Delta {u^\sharp }(x) = {| {{x^\sharp }} |^{n + 2}}(\Delta u)\left( {{x^\sharp }} \right) = \frac {1} {{{{\left| x \right|}^{n + 2}}}}(\Delta u)\left( {\frac {x} {{{{\left| x \right|}^2}}}} \right)$.

We now consider a different situation. The detail can be found in the following paper due to X.W. Xu published in Proc. Roy. Soc. Edinburgh Sect. A, 2000.

Theorem. If $u$ is a sufficiently good function then $v$ satisfies the equation

$\displaystyle \Delta^2 v = \frac{1}{|x|^{n+4}}(\Delta^2 u)\left(\frac{x}{|x|^2}\right)$

where $v$ is defined to be

$\displaystyle v(x)=|x|^{4-n}u\left(\frac{x}{|x|^2}\right)$.